Problem: The equation of a circle $C$ is $x^2+y^2-8x-2y-8 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Explanation: To find the equation in standard form, complete the square. $(x^2-8x) + (y^2-2y) = 8$ $(x^2-8x+16) + (y^2-2y+1) = 8 + 16 + 1$ $(x-4)^{2} + (y-1)^{2} = 25 = 5^2$ Thus, $(h, k) = (4, 1)$ and $r = 5$.